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I'm having some trouble with the following exercise:

Let $R$ be a Noetherian commutative ring, $I$ and ideal and $J=\bigcap_{n\geq0}I^n$. Show that $IJ = J$. (Hint: Assume that $J\not\subseteq IJ$ and consider a primary decomposition of $IJ$. Take $x\in J\setminus Q$ for some $Q$ primary component of $IJ$).

So, let's assume that $J\not\subseteq IJ$ and let $x\in J\setminus Q$ where $Q$ is in the primary decomposition of $IJ$. If $Q$ were prime instead of primitive, then $J\setminus Q$ would be multiplicatively closed, and thus $x^n\in J\setminus Q$ would give rise to a contradiction because $x^n\in IJ\subseteq Q$. But $Q$ being primitive we can't conclude this.

I tried to look at the proof that $Q$ prime $\implies$ $J\setminus Q$ multiplicatively closed, tried to adapt it with $Q$ being primitive, and arrived at the following: If $x\in J\setminus Q$ and $\forall n \in \mathbb N, y^n\in J\setminus Q$, then $xy \in J\setminus Q$. But now I don't know how to continue from here. If we prove some power of $x$ is in $J\setminus Q$ then we find a contradiction, and it feels like this way I'm getting very close to that goal.

What are your thoughts about this?

656475
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  • Given said $x$, show that for all $i\in I$, there is some $n\in \mathbb N$ such that $i^n\in Q$, then use Noetherianity to show that there is some $k$ such that $I^k\subseteq Q$ which implies $J\subseteq Q$ – Mor A. Jan 04 '24 at 17:38
  • I was able to prove that for all $i\in I$ there is some $n$ such that $i^n\in Q$, but how do I use noetherianity to conclude that $I^k\subseteq Q$ for some k? @MorA. – 656475 Jan 04 '24 at 18:04
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    https://math.stackexchange.com/questions/4838114 – Martin Brandenburg Jan 04 '24 at 18:22

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